In how many ways four “+” and five “-“can be arranged in a circle so that no two “+” sign are together?
 $
  (a){\text{ 4!}}{\text{.5!}} \\
  (b){\text{ 2}} \\
  (c){\text{ 4!}}{\text{.4!}} \\
  (d){\text{ 1}} \\
$

Hint – In this question we have to place some “+” signs and some “-“signs into a circle, so this problem resembles somewhat of a round table arrangement. If we have to place n entities into a round table then there can be $(n - 1)!$ ways to do it so. Use this concept to get the answer.

Complete step by step answer:
As we know that four people cannot sit between three people.
So we have to first arrange five negative (-) signs on a round table.
As we know if there are x candidates arrange on a round table the number of ways is $\left( {x - 1} \right)!$
Therefore the number of ways to arrange five negative (-) signs on a round table is $\left( {5 - 1} \right)!$ = (4!).
Now we have to arrange four positive (+) signs between them so that no two (+) signs are together.
So as we know between five negative (-) signs there are five vacant spaces therefore the number of ways to arrange four (+) signs between five (-) signs so that no two (+) signs are together is (5!).
Therefore the total number of ways to arrange five (-) and four (+) sign so that no two (+) sign are together is
$ \Rightarrow \left( {4!} \right)\left( {5!} \right)$
So, this is the required answer.
Hence option (A) is correct.

Note – Whenever we face such types of problems the key concept that needs to come up into mind is that of the possible arrangements possible into a round table arrangements. Don’t forget to rearrange these signs amongst themselves using the concept of permutation, this will help you get on the right track to get the answer.